Josia Pietsch
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341 lines
13 KiB
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341 lines
13 KiB
TeX
\lecture{14}{2023-12-04}{}
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\gist{%
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\begin{abuse}
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Sometimes we say club
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instead of club in $\kappa$.
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\end{abuse}
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\begin{example}
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Let $\kappa$ be a regular uncountable cardinal.
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\begin{itemize}
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\item $\kappa$ is a club in $\kappa$.
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\item $\{\xi + 1 : \xi < \kappa\}$ is unbounded in $\kappa$,
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but not closed.
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\item For each $\alpha < \kappa$,
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the set $\alpha + 1 = \{\xi : \xi \le \alpha\}$
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is closed but not unbounded in $\kappa$.
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\item $\{\xi < \kappa : \xi \text{ is a limit ordinal}\} $
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is club in $\kappa$.
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\end{itemize}
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\end{example}
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}{}
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\begin{lemma}
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\label{lem:clubintersection}
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Let $\kappa$ be regular and uncountable.
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Let $\alpha < \kappa$
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and let $\langle C_{\beta} : \beta < \alpha \rangle$
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be a sequence of subsets of $\kappa$ which are all club in $\kappa$.
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Then
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\[
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\bigcap_{\beta < \alpha} C_{\beta}
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\]
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is club in $\kappa$.
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\end{lemma}
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\begin{warning}
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This is false for $\alpha = \kappa$:
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Let $C_{\beta} \coloneqq \{\xi : \xi > \beta\}$.
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Clearly this is club
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but $\bigcap_{\beta < \kappa} C_\beta = \emptyset$.
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\end{warning}
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\begin{yarefproof}{lem:clubintersection}
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\gist{%
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First let $\alpha = 2$.
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Let $C, D \subseteq \kappa$
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be club.
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$C \cap D$ is trivially closed:
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Let $\beta < \kappa$. Suppose that $(C \cap D) \cap \beta$
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is unbounded in $\beta$, so $C \cap \beta$ and $D \cap \beta$
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are both unbounded in $\beta$,
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so $\beta \in C \cap D$.
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$C \cap D$ is unbounded:
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Take some $\gamma < \kappa$.
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Let $\gamma_0 = \gamma$
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and inductively define $\gamma_n$ :
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If $n$ is even, let $\gamma_n \coloneqq \min C \setminus (\gamma_{n-1}+1)$,
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otherwise $\gamma_n \coloneqq \min D \setminus (\gamma_{n-1}+1)$.
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Let $\xi = \sup \{\gamma_n : n < \omega\}$.
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Then $\xi = \sup \{\gamma_{2n + 2} : n < \omega\} \in D$
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and $\xi \in C$ by the same argument,
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so $\xi \in C \cap D$
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(here it is important, that $\cf(\kappa) > \omega$)
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and $\xi > \gamma$.
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The case $\alpha > 2$ is similar:
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The intersection is closed by exactly the same argument.%
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\footnote{``It is even more closed.''}
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Let's prove that $\bigcap \{C_{\beta} : \beta < \alpha\}$
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is unbounded in $\kappa$.
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We will define a sequence $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$%
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\footnote{Ordinal multiplication, i.e.~$\alpha \cdot \omega = \sup_{n < \omega} \underbrace{\alpha + \ldots + \alpha}_{n \text{ times}}$.}
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as follows:
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Let $\gamma_0 \coloneqq \gamma$.
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Choose
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\[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\]
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and at limits choose the supremum.
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Let $\xi = \sup_{i < \alpha \cdot \omega} \gamma_i
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= \sup_{i < \omega} \gamma_{\alpha \cdot n + \beta + 1} \in \bigcap_{\beta < \alpha} C_\beta$,
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where we have used that.
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$\cf(\kappa) > \alpha \cdot \omega$.
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}{%
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\begin{itemize}
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\item Trivially closed.
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\item Recursively define $\langle \gamma_i : i \le \alpha \cdot \omega \rangle$, by
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$\gamma_0 \coloneqq \gamma$,
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\[\gamma_{\alpha \cdot n + \beta + 1} = \min C_{\beta} \setminus (\gamma_{\alpha \cdot n + \beta} + 1)\]
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and $\sup$ at limits.
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\item Then $\sup_{i < \alpha\cdot \omega} \gamma_i \in \bigcap_{\beta < \alpha} C_\beta$
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(we used $\cf(\kappa) > \alpha\cdot \omega$).
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\end{itemize}
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}
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\end{yarefproof}
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\begin{definition}
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$F \subseteq \cP(a)$ is a \vocab{filter}
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iff
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\begin{enumerate}[(a)]
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\item $X,Y \in F \implies X \cap Y \in F$,
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\item $X \in F \land X \subseteq Y \subseteq \kappa \implies Y \in F$,
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\item $\emptyset \not\in F$,\footnote{Some authors don't
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require $\emptyset \not\in F$,
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but that is a degenerate case anyway,
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since $\emptyset \in F \iff F = \cP(a)$.}
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$\kappa \in F$.
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\end{enumerate}
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Let $\alpha \le \kappa$.
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We call $F$ \vocab{$< \alpha$-closed}
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iff for all $\gamma < \alpha$ and $\{X_\beta : \beta < \gamma\} \subseteq F$
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then $\bigcap \{X_\beta : \beta < \gamma\} \in F$.
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\end{definition}
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\gist{%
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Intuitively, a filter is a collection of ``big'' subsets of $a$.
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}{}
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\begin{definition}
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Let $\kappa$ be regular and uncountable.
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The \vocab{club filter} is defined as
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\[
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\cF_{\kappa} \coloneqq \{X \subseteq \kappa : \exists \text{ club } C \subseteq \kappa .~ C \subseteq X\}.
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\]
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\end{definition}
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Clearly this is a filter.
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We have shown (assuming \AxC to choose contained clubs):
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\begin{theorem}
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If $\kappa$ is regular and uncountable.
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Then $\cF_\kappa$ is a $< \kappa$-closed filter.
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\end{theorem}
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\begin{proof}
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Clearly $\emptyset \not\in \cF_\kappa$,
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$\kappa \in \cF_\kappa$,
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and $A \in \cF_{\kappa}, A \subseteq B \in \kappa \implies B \in \cF_\kappa$.
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In \autoref{lem:clubintersection} showed that the intersection
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of $< \kappa$ many clubs is club.
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\end{proof}
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\begin{definition}
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Let $\langle A_\beta : \beta < \alpha \rangle$
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be a sequence of sets.
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The \vocab{diagonal intersection},
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is defined to be
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\[
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\diagi_{\beta < \alpha} A_{\beta} \coloneqq
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\{\xi < \alpha : \xi \in \bigcap \{A_{\beta} : \beta < \xi\} \}
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= \bigcap_{\beta < \alpha} ([0,\beta] \cup A_\beta)
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\]
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\end{definition}
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\begin{remark}+
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\label{rem:diagiclosed}
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Note that if $A$ is closed,
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so is $[0,\alpha] \cup A$.
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Since the intersection of arbitrarily many
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closed sets is closed,
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we get that the diagonal intersection
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of closed sets is closed.
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\end{remark}
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\begin{lemma}
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\label{lem:diagiclub}
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Let $\kappa$ be a regular, uncountable cardinal.
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If $\langle C_{\beta} : \beta < \kappa \rangle$
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is a sequence of club subsets of $\kappa$,
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then $\diagi_{\beta < \kappa} C_{\beta}$
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contains a club.
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\end{lemma}
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\begin{yarefproof}{lem:diagiclub}
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% TODO THINK
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\gist{
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Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$.
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Write $D_{\beta} \coloneqq \bigcap \{C_{\gamma} : \gamma \le \beta\} $
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for $\beta < \kappa$.
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Each $D_{\beta}$ is a club,
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$D_{\beta} \subseteq C_{\beta}$
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and $D_{\beta} \supseteq D_{\beta'}$
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for $\beta \le \beta' < \kappa$.
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It suffices to show that $\diagi_{\beta < \kappa} D_{\beta}$
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contains a club.
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\begin{claim}
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$\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$.
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\end{claim}
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\begin{subproof}
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Cf.~\yaref{rem:diagiclosed}.
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% Let $\gamma < \kappa$ be such that $\left( \diagi_{\beta < \kappa} D_{\beta} \right) \cap \gamma$
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% is unbounded in $\gamma$.
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% We aim to show that $\gamma \in \diagi_{\beta < \kappa} D_{\beta}$.
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% Let $\beta_0 < \gamma$.
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% We need to see that $\gamma \in D_{\beta_0}$.
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% For each $\beta_0 \le \beta' < \gamma$
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% there is some $\beta'' \in \diagi_{\beta < \kappa} D_\beta$
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% such that $\beta' \le \beta'' < \gamma$,
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% since $\gamma = \sup((\diagi_{\beta < \kappa} D_\beta) \cap \gamma)$.
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% In particular $\beta'' \in D_{\beta_0}$.
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% So $D_{\beta_0} \cap \gamma$
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% is unbounded in $\gamma$.
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% Since $D_{\beta_0}$ is closed
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% it follows that $\gamma \in D_{\beta_0}$.
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%As $\beta_0 < \gamma$ was arbitrary,
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%this shows that $\gamma \in \diagi_{\beta < n} D_\beta$.
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\end{subproof}
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\begin{claim}
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$\diagi_{\beta < \kappa} D_{\beta}$
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is unbounded in $\kappa$.
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\end{claim}
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\begin{subproof}
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Fix $\gamma < \kappa$.
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We need to find $\delta > \gamma$
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with $\delta \in \diagi_{\beta < \kappa} D_\beta$.
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Define $\langle \gamma_n : n < \omega \rangle$
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as follows:
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$\gamma_0 \coloneqq \gamma$
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and
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\[
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\gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1)
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\]
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We have $\delta \coloneqq \sup_{n < \omega} \gamma_n \in \kappa$
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by cofinality of $\kappa$.
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We need to show that $\delta \in D_{\overline{\gamma}}$
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for all $\overline{\gamma} < \delta$.
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If $\overline{\gamma} < \delta$, then $\overline{\gamma} \le \gamma_n$
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for some $n < \omega$.
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For $m \ge n$, $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_{\overline{\gamma}}$.
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So $D_{\overline{\gamma}} \cap \delta$ is unbounded
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in $\delta$, hence $\delta \in D_{\overline{\gamma}}$.
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\end{subproof}
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}{%
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\begin{itemize}
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\item Fix $\langle C_\beta : \beta < \alpha \rangle$.
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Set $D_{\beta} \coloneqq \bigcap_{\gamma \le \beta} D_{\gamma}$.
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It suffices to analyze $D_{\beta}$.
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\item $\diagi_{\beta < \kappa} D_{\beta}$ is closed in $\kappa$:
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\begin{itemize}
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\item Let $\gamma < \kappa$ such that $\left( \diagi_{\beta < \kappa} D_{\beta}\right) \cap \gamma$ unbounded in $\gamma$.
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Want $\gamma \in \diagi_{\beta < \kappa} D_\beta$.
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\item Let $\beta_0 < \gamma$. Want $\gamma \in D_{\beta_0}$.
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\item $D_{\beta_0} \cap \gamma$ is unbounded in $\gamma$
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($D_{\beta_0} \setminus \beta_0 \supseteq \diagi_{\beta < \kappa} D_{\beta} \setminus \beta_0$)
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$\overset{D_{\beta_0} \text{ closed}}{\implies} \gamma \in D_{\beta_0}$.
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\end{itemize}
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\item $\diagi_{\beta < \kappa} D_\beta$ is unbounded in $\kappa$:
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\begin{itemize}
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\item Fix $\gamma < \kappa$. We need to find $\diagi_{\beta < \kappa} D_\beta \ni \delta > \gamma$.
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\item Define $\langle \gamma_n : n < \omega \rangle$
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by $\gamma_0 \coloneqq \gamma$,
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$\gamma_{n+1} \coloneqq \min D_{\gamma_n} \setminus (\gamma_n + 1)$,
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$\delta \coloneqq \sup_n \gamma_n \overset{\cf(\kappa) > \omega}{<} \kappa$
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\item Want $\delta \in \diagi_{\beta < \kappa} D_\beta$,
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i.e.~$\forall \epsilon < \delta.~\delta\in D_\epsilon$.
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If $\epsilon < \delta$, then $\epsilon \le \gamma_n$
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for $n$ large enough,
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so $\gamma_{m+1} \in D_{\gamma_m} \subseteq D_{\gamma_n} \subseteq D_\epsilon$
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for $m \ge n$.
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Thus $\sup(D_\epsilon \cap \delta) = \delta$
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$\overset{D_\epsilon \text{ closed}}{\implies} \delta \in D_{\epsilon}$.
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\end{itemize}
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\end{itemize}
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}
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\end{yarefproof}
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\begin{remark}+
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$\diagi_{\beta < \kappa} C_{\beta}$ actually
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\emph{is} a club,
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since $\diagi_{\beta < \kappa} C_\beta$ is closed,
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again cf.~\yaref{rem:diagiclosed}.
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% Let $\lambda < \kappa$ be a limit ordinal.
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% Suppose that $\lambda \not\in \diagi_{\beta < \kappa} D_\beta$.
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% Then there exists $\alpha < \lambda$ such that
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% $\lambda \not\in D_\alpha$.
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% Since $D_\alpha$ is closed,
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% we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
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% In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
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\end{remark}
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\begin{definition}
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Let $\kappa$ be regular and uncountable.
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$S \subseteq \kappa$ is called \vocab{stationary} (in $\kappa$)
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iff $C \cap S \neq \emptyset$
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for every club $C \subseteq \kappa$.
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\end{definition}
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\begin{remark}+[\url{https://mathoverflow.net/q/37503}]
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Informally, club sets and stationary sets
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can be viewed as large sets of a measure space
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of measure $1$.
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Clubs behave similarly to sets of measure $1$
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and stationary sets are analogous to
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sets of positive measure:
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\begin{itemize}
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\item Every club is stationary,
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\item the intersection of two clubs is a club,
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\item the intersection of a club and a stationary set is stationary,
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\item there exist disjoint stationary sets.
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\end{itemize}
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\end{remark}
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\begin{example}
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\begin{itemize}
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\item Every $D \subseteq \kappa$ which is club in $\kappa$
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is stationary in $\kappa$.
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\item There exist disjoint stationary sets:\footnote{Note that clubs can never be disjoint, since their intersection is a club.}
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Let $\kappa = \omega_2$.
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Let $S_0 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega\}$
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and $S_1 \coloneqq \{\xi < \kappa : \cf(\xi) = \omega_1\}$.
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Clearly these are disjoint.
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They are both stationary:
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Let $C \subseteq \kappa$ be a club.
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Let $(\xi_i : i \le \omega_1)$
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be defined as follows:
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$\xi_0 \coloneqq \min C$,
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$\xi_i \coloneqq \min (C \setminus \sup_{j < i} \xi_j)$.
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For $i \le \omega_1$ we have that $\xi_i = \sup_{j < i} \xi_j$.
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In particular $\xi_\omega \in S_0 \cap C$
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and $\xi_{\omega_1} \in S_1 \cap C$.
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\end{itemize}
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\end{example}
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We will show later that if $ \kappa$ is a regular uncountable cardinal,
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then every stationary $S \subseteq \kappa$ can be written as
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$S = \bigcup_{i < \kappa} S_i$,
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where the $S_i$ are stationary and pairwise disjoint.
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